The two boxcars A and B have a weight of 20 000 Ib and 30 000 Ib, respectively. If they coast freely down the incline when the brakes are applied to all the wheels of car A causing it to skid, determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is μk=0.5. The wheels of car B are free to roll. Neglect their mass in calculation.

a = 3.608 ft/s²

T = 5.977 kip

Explanation:

Angle of inclined track is not given. It is 5 degree as shown in attached image 1.

Given data:

Weight of boxcar A = Wα = 20,000 lb

Weight of boxcar B = Wb = 30,000 lb

Coefficient of kinetic friction = υk = 0.5

angle of inclined track = 5 degree

Value of g = 32.2 ft/s²

Solution:

These boxcars are moving therefore,

∑Fₓ = ma

∑

Car A

Use ∑

Forces acting in Y-axis are weight and its counter N (as shown in Free body diagram). Weight is Negative (acting downward) while N is positive (acting upward)

∑ Nα - Wα*Cos(∅) = 0

∑ Nα - 20,000*Cos(5°) = 0

∑ Nα - 20,000*0.9962 = 0

∑ Nα - 19,924 = 0                

Nα = 19,924

Use ∑Fₓ = maₓ

Forces acting in x-axis are Fa, T and horizontal component of weight and force having acceleration a (as shown in Free body diagram). Horizontal component of weight and Tension (T) are Negative (acting in negative X-axis direction) while Fa is positive (acting in positive X-axis direction)

∑Fₓ = maₓ

(0.5*19924) - 20,000*Sin(5°) - T = (20,000/32.2)*a

9962 - 1744 - T = 621.12*a

8218 - T = 621.12*a  .................. Eq(1)

Both cars

∑Fₓ = maₓ

Forces acting are; horizontal component of weights, Fa and force having acceleration a (as shown in Free body diagram).

9962 - 1744 - 2616 = 621.12/a + 931.68/a

5602 = 1552.8 / a

⇒ a = 3.608 ft/s²

put a = 3.608 ft/s² in equation (1)

⇒ T = 8218 - 621.12*3.608

    T = 8218 - 2240.79

    T = 5977.21 lb

    T = 5.977 kip